Find $\lim_{x\to0} \cos\left(2\pi x \right)^{^{\frac 1 x}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $0$ (Choice C) C $e^2$ (Choice D) D The limit doesn't exist.
Substituting $x=0$ into $\cos\left(2\pi x \right)^{^{\frac 1 x}}$ results in the indeterminate form $1^{^{\infty}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=\cos\left(2\pi x \right)^{\frac 1 x}$, we will find $\lim_{x\to 0}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to 0}y$. $\ln(y) =\dfrac{\ln\left[\cos\left(2\pi x \right)\right]}{x}$ Substituting $x=0$ into $\dfrac{\ln\left[\cos\left(2\pi x \right)\right]}{x}$ results in the indeterminate form $\dfrac{0}{0}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to 0}\ln(y) \\\\ &=\lim_{x\to 0}\dfrac{\ln\left[\cos\left(2\pi x \right)\right]}{x} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}(\ln[\cos\left(2\pi x \right)])}{\dfrac{d}{dx}[x]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{-2\pi\sin(2\pi x)}{\cos\left(2\pi x \right)}}{1} \\\\ &=\dfrac{0}{1} \gray{\text{Substitution}} \\\\ &=0 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\ln\left[\cos\left(2\pi x \right)\right]}{x}$ actually exists. We found that $\lim_{x\to 0}\ln(y)=0$, which means $\lim_{x\to 0}y=1$. [Why?] In conclusion, $\lim_{x\to 0}\dfrac{\ln\left[\cos\left(2\pi x \right)\right]}{x}=1$.